# Answer to Question #11133 in Physical Chemistry for Emma Van Sickle

Question #11133
OK so the equation is balanced and is:
3Mg + 2AlCl(3) ---&gt; 3MgCl(2) + 2Al
I am given:
mass of crucible and Mg: 34.61g
mass of crucible alone: 28.43g
mass of crucible and Al: 32.42g

I was told to subtract the crucible weight from the other 2 measurements so Mg = 6.18g and Al = 3.99g

IThe questions are:
1) What mass of Mg is used in the reaction?
2) What is the mass of Al produced?
3) What is the Theoretical Yield of AL?
4) What is the percent yield of the reaction?
1
2012-06-22T08:38:43-0400
1) m(Mg) = m(crucible and Mg) - m(crucible alone) = 34.61 - 28.43 = 6.18 g
2)
m(Al) = m(crucible and Al) - m(crucible alone) = 32.42 - 28.43 = 3.99 g
3)
m(Al, theor.) = m(Mg)*2*M(Al) / 3*M(Mg) = 6.18*2*27 / (24*3) = 4.635 g
4) the
percent yield is (3.99 / 4.635)*100% = 86.1%

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!